mjhjmtu

I want to put a bunch of executable scripts in the .command dir (which is also executable), and then only have to source that directory in my .bash_profile. Is this possible? I can get this to work with one file. But when adding a second file, the second file's commands aren't available in the shell.

my .bashprofile

source ~/.commands/*

my .commands folder

-rwxr-xr-x 1 christopherreece staff 108 Dec 14 08:55 server_utils.sh
-rwxr-xr-x 1 christopherreece staff 23 Dec 14 09:04 short

contents of short

echo 'a short program'

contests of server_utils.sh

function upfile 
 scp $1 root@myserveripadress:~/

Shell input and output.

$ hello
hello

$ short
-bash: short: command not found

You can't do that with one source. The first argument is taken as the file name, the others show up as the positional parameters $1, $2... in the sourced script.

$ cat test.src 
echo hello $1
$ source test.src there
hello there

But you could do it with a loop:

for f in ~/commands/*.src; do
 source "$f"
done

_{(As an aside, having stuff like that include only files with a certain extension is quite useful if you use an editor that leaves backup files with a trailing ~. The backup copies don't become accidentally active, then.)}

Though note that if you have a sourced script that contains plain commands (like that echo above or your short), they'll be executed when the script is sourced. They don't generate any functions in the sourcing shell.

$ cat test2.src 
echo "shows when sourced"
func() 
 echo "shows when function used"
 
$ source test2.src 
shows when sourced
$ func
shows when function used

If you want to have executable scripts instead, the kind where the script runs when you give its name as a command, put them somewhere in PATH (I'd suggest using ~/bin for that), give them the execute permission and put proper hashbangs in the beginning of the scripts (#!/bin/sh or #!/bin/bash or whatever)